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Problem 556

Posted: Sun Apr 17, 2016 1:31 pm
by v6ph1
Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)

Re: Problem 556

Posted: Sun Apr 17, 2016 3:28 pm
by mpiotte
v6ph1 wrote:Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
Thanks, corrected.

Re: Problem 556

Posted: Mon Apr 08, 2024 6:28 pm
by Swistakk
mpiotte wrote: Sun Apr 17, 2016 3:28 pm
v6ph1 wrote:Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
Thanks, corrected.
Well, not really...

Re: Problem 556

Posted: Sat Jul 27, 2024 7:32 pm
by echip
mpiotte wrote: Sun Apr 17, 2016 3:28 pm
v6ph1 wrote:Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
Thanks, corrected.
You say it's corrected, But I see (1+i)*(1+2*i)

Re: Problem 556

Posted: Sun Jul 28, 2024 9:22 am
by skoczian
mpiotte wrote: Sun Apr 17, 2016 3:28 pm
v6ph1 wrote:Problem 556 (View Problem)
The Factorization of 1+3i seams to be wrong:
It should be (1+i)*(2+i)
Thanks, corrected.
This factorization occurs twice in the problem text. Only once it's corrected:
A Gaussian integer can be uniquely factored as the product of a unit and proper Gaussian primes.
For example 2 = -i(1 + i)2 and 1 + 3i = (1 + i)(2 + i).
Here it's correct.
And in the next paragraph:
For example f(10) = 7 because 1, 1 + i, 1 + 2i, 1 + 3i = (1 + i)(1 + 2i), ...
Still wrong.

Re: Problem 556

Posted: Sun Jul 28, 2024 12:12 pm
by RobertStanforth
Thank you everyone for flagging this. The second occurrence has now been corrected to read (1+3i)=(1+i)(2+i).